## AI-GEOSTATS: Calculating averages

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• Hello, I was hoping advice could be given on the following problem. We store nickel (eg 0.40 Ni %) and mill recovery (55.45%) in our mine model. From this we
Message 1 of 3 , Jan 5, 2003
Hello,

I was hoping advice could be given on the following problem. We store nickel
(eg 0.40 Ni %) and mill recovery (55.45%) in our mine model. From this we
calculate a mill recovered nickel (eg 0.22 RecNi %) and store this in the
model. All 3 model items are stored to two decimal places. The rounding is
to three decimal places (i.e. +/-0.005).

We create ore-cuts based on the tonnes-weighted average of the Mill
Recovered Nickel (using an Ore-material classification scheme based off Mill
Recovered Nickel).

My problem is given below and highlighted in green. I have assumed tonnage
weighting in each row is 1.

Nickel Mill Recovered
0.53 60 0.3180
0.65 62 0.4030
0.32 20 0.0640
0.8 70 0.5600
0.56 63 0.3528
0.45 55 0.2475
0.58 68 0.3944
Average 0.555714286 56.85714286 0.3342

divided by Average Mill Recovery ====> 0.315963265

It would appear that the averaging of the recovered nickel (0.3342) gives a
different result to dividing the average Nickel Head Grade by the Average
Mill Recovery (0.315963265). These averages are generated in Microsoft Excel
so it doesn't seem to be a rounding error. We believe the first average is
wrong and it would appear that the greater the variability of the data set,
the greater the difference between the two calculation averages. If
possible, would anyone know what is happening here and which is the correct
approach.

Thank-you in appreciation.

Craig Allison

MKO Resource Evaluation Geologist

[Non-text portions of this message have been removed]
• Calculating averagesCraig, Don t you mean multiply Average Nickel Head grade by Average Mill recovery. Your first figure 0.3342 is correct but your figure
Message 2 of 3 , Jan 5, 2003
Calculating averagesCraig,
is incorrect. It is your assumption that your average mill recovery is
tonneage recovery, i.e. you averaged the mill recovery as if it where
recovery per tonne of ore instead of per % of metal so your calculation
of 0.5557*0.56857 is incorrect.

Your average mill recovery should be calculated

(0.53*60)+(0.65*62)+(0.32*20)+(0.8*70)+(0.56*63)+(0.45*55)+(0.58*0.68)/
(0.53+0.65+0.32+0.8+0.56+0.45+0.58) =0.601465

So your original table should look like;

Nickel Mill Recovered
0.53 60 0.3180
0.65 62 0.4030
0.32 20 0.0640
0.8 70 0.5600
0.56 63 0.3528
0.45 55 0.2475
0.58 68 0.3944
Average 0.555714286 60.14652956 0.3342

divided by Average Mill Recovery ====> 0.334242857

Regards Digby Millikan
Geolite Mining Systems
digbym@...
http://www.users.on.net/digbym

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• Hi Craig The average of a product will only equal the product of the averages if the two numbers are completely independent of one another. This is exactly
Message 3 of 3 , Jan 5, 2003
Hi Craig

The average of a product will only equal the product
of the averages if the two numbers are completely
independent of one another.

This is exactly analogous to the calculation of a
covariance in statistics.

Isobel
http://uk.geocities.com/drisobelclark

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